Page 142 - Essentials of applied mathematics for scientists and engineers
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book   Mobk070    March 22, 2007  11:7








                     132  ESSENTIALS OF APPLIED MATHEMATICS FOR SCIENTISTS AND ENGINEERS
                            We now use the convolution principle to evaluate the solution for the general case of
                       q(τ). We are searching for the inverse transform of
                                                      √                      √
                                              1 cosh(ς s )   Q(s )     cosh(ς s )
                                                     √    +        1 −       √
                                              s cosh   s      s         cosh  s
                            The inverse transform of the first term is given above. As for the second term, the inverse
                       transform of Q(s )issimply q(τ) and the inverse transform of the second term, absent Q(s) is

                                                   2(−1) n+1  cos    2n−1 ςπ     2n−1 2  2
                                               1 −               2      e −(  2 ) π τ
                                                         (2n − 1)π
                            According to the convolution principle, and summing over all poles

                                        ∞
                                         2(−1)  n+1  cos     2n−1 ςπ     2n−1 2  2
                              u(ς, τ) =                  2     e −(  2 ) π τ
                                                (2n − 1)π
                                       n=1

                                          ∞    τ            n+1     2n−1
                                                       2(−1)    cos  2  ςπ    −(  2n−1 2  2

                                       +           1 −                       e    2 ) π τ  q(τ − τ )dτ
                                              τ =0           (2n − 1)π

                                          n=1
                       Example 8.6. Next consider heat conduction in a semiinfinite region x > 0, t > 0. The initial
                       temperature is zero and the wall is subjected to a temperature u(0, t) = f (t)atthe x = 0
                       surface.
                                                           u t = u xx
                                                           u(x, 0) = 0
                                                           u(0, t) = f (t)

                       and u is bounded.
                            Taking the Laplace transform and applying the initial condition

                                                            sU = U xx

                       Thus
                                                                 √            √
                                                U(x, s ) = A sinh x s + B cosh x s
                            Both functions are unbounded for x →∞. Thus it is more convenient to use the
                       equivalent solution
                                                             √       √         √
                                               U(x, s ) = Ae  −x s  + Be x s  = Ae  −x s

                       in order for the function to be bounded. Applying the boundary condition at x = 0

                                                            F(s ) = A
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